Certainly! Let's factor the given polynomial over the integers: \( x^2 - 6x - 16 \).
1. Identify the quadratic polynomial:
[tex]\[ x^2 - 6x - 16 \][/tex]
2. Set up the expression to be factored:
We need to find two binomials of the form:
[tex]\[ (x + a)(x + b) \][/tex]
such that when you expand the product, you get back the original polynomial:
[tex]\[ x^2 - 6x - 16 = x^2 + (a+b)x + ab \][/tex]
3. Determine the values of \( a \) and \( b \):
- We need the coefficients \( a \) and \( b \) to satisfy two conditions:
1. The sum of \( a \) and \( b \) must equal the coefficient of the linear term (which is \(-6\)):
[tex]\[ a + b = -6 \][/tex]
2. The product of \( a \) and \( b \) must equal the constant term (which is \(-16\)):
[tex]\[ ab = -16 \][/tex]
4. Find pairs (a, b) that satisfy these conditions:
Let's find pairs of integers whose product is \(-16\):
[tex]\[
(-1, 16), (1, -16), (-2, 8), (2, -8), (-4, 4), (4, -4)
\][/tex]
Now, let's find the pair that also adds up to \(-6\):
- \((-8, 2)\):
[tex]\[
(-8) + 2 = -6 \\
(-8) \cdot 2 = -16
\][/tex]
This pair satisfies both conditions.
5. Write the factored form:
- With \( a = -8 \) and \( b = 2 \):
[tex]\[ x^2 - 6x - 16 = (x - 8)(x + 2) \][/tex]
6. Verification:
- Expanding \( (x - 8)(x + 2) \):
[tex]\[
(x - 8)(x + 2) = x^2 + 2x - 8x - 16 = x^2 - 6x - 16
\][/tex]
The factored form is correct.
Thus, the factored form of \( x^2 - 6x - 16 \) over the integers is:
[tex]\[ (x - 8)(x + 2) \][/tex]
